Tag Info

7

It is possible to get a number of distinct coefficients exponential in $n$. Here is an example. Let B = \begin{bmatrix} 1 & -1 & -1 & -1 & \cdots & x_1 \\ 0 & 1 & 0 & 0 & & x_2 \\ 0 & 0 & 1 & 0 & \cdots & x_3 \\ 0 & 0 & 0 & 1 & & \vdots \\ & \vdots & & & \...

3

$\newcommand\Z{\mathbb Z}$Point 1 has already been answered by @spin, and by @nfdc23 at Centralizers of subtori in reductive groups, derived subgroups. In light of point 1, point 2 is the same as asking whether there are a group $G$ with derived group $G' = \operatorname{SL}_2$, a maximal torus $T$ in $G$, and a root $\alpha$ of $T$ in $\operatorname{Lie}(G)$...

3

I think looking at Steinberg's lecture notes might be instructive for this question. Following the notation $h_{\alpha}(t)$ defined in Steinberg's lecture notes (Lemma 19), the homomorphism $\operatorname{SL}_2 \rightarrow G$ corresponding to a root $\alpha$ has image $\operatorname{PGL}_2$ if and only if $h_{\alpha}(-1) = 1$. By Lemma 28 (c) in Steinberg, ...

2

Having a bound on $\|M\|_1$, you get by duality a bound on $\|M\|_\infty$ and then you get a bound on $\|M\|_p$ for every $1\leq p\leq\infty$ by the Riesz–Thorin theorem.

1

You can find a version of Maschke's Theorem for group rings over arbitrary coefficient rings with 1, commutative or not, in Milies & Sehgal's An Introduction to Group Rings. See Theorem 3.4.7. It states that $R[G]$ is semisimple if, and only if: (1) $R$ is a semisimple ring; (2) $G$ is a finite group; (3) $|G|$ is a unit in $R$.

1

An answer is given in some sense in https://arxiv.org/abs/1907.06552, the first paragraph in p.3. A quiver variety M associated with a nonsymmetric Cartan matrix does not make sense, but the space of maps from P^1 to a hypothetical M makes sense.

1

Lassalle and Schlosser have obtained in Inversion of the Pieri formula for Macdonald polynomials some recurrence relations for MacDonald polynomials $P_\lambda(x;q,t)$ which can be restricted to Jack polynomials making $q=t^\alpha$ and letting $t\to 1$. These recurrence relations lead to expressions for Jack polynomials as linear combination of products of ...

Only top voted, non community-wiki answers of a minimum length are eligible